Question

The shortest pipe in a particular organ is 1.23 m.

Answer 1

Answer:

The 9th harmonic is 1,010.67 Hz.

Explanation:

Given;

length of the pipe, L = 1.23 m

speed of sound at 0°C = 331.5 m/s

A pipe closed at one end is known as a closed pipe;

The wavelength of the sound for the first harmonic is calculated as;

L = Node ------ > Antinode

$L = \frac{\lambda }{4} \\\\\lambda = 4L$

First harmonic:  $F_0 = \frac{V}{\lambda} = \frac{V}{4L}$

The wavelength of the sound for the first harmonic is calculated;

L = Node ----> Node + Node ------> Antinode

$L = \frac{\lambda}{2} + \frac{\lambda}{4} = \frac{2\lambda+\lambda}{4} = \frac{3\lambda}{4} \\\\\lambda = \frac{4 L}{3}$

Second harmonic: $F_1 = \frac{V}{\lambda} = \frac{3V}{4L}$

F₁ = 3F₀

The increment from F₀ to F₁ is 1 to 3; (odd number).

(1, 3, 5, 7, 9, 11, 13, 15, 17, 19..... n+2, where n is odd number)

The 9th harmonic = F₈

F₈ = 15F₀

$F_8= 17(F_0) = 15 (\frac{V}{4L} )\\\\F_8 = 15(\frac{331.5}{4\times 1.23} )\\\\F_8 = 1,010.67 \ Hz$

Therefore, the 9th harmonic is 1,010.67 Hz.